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Lattice reduction

Overview

Having introduced the LLL reduction, we now provide a more general notions of a reduced basis for a lattice as well as provide bounds for the basis vectors. The key idea behind introducing these definitions is that once we know some basis vector is []-reduced, we can bound the sizes of the basis, which is important when algorithms require short vectors in a lattice. For fast algorithms, LLL-reduction is typically the most important notion as it can be computed quickly. Two main definitions appear often when discussing lattice reductions, which we will provide here.

Definitions

A basis{bi}i=1d\left\{b_i\right\}_{i=1}^d{bi​}i=1d​is size-reduced if ∣μi,j∣≤12\left|\mu_{i,j}\right|\leq\frac12∣μi,j​∣≤21​. Intuitively this captures the idea that a reduced basis being "almost orthogonal".

Let LLLbe a lattice, 1≤i≤dim⁡L=d1\leq i\leq\dim L=d1≤i≤dimL=d, we define the ithi^\text{th}ithsuccessive minimaλi(L)\lambda_i(L)λi​(L) as

λi(L)=min⁡(max⁡1≤j≤i(∥vj∥):vj∈L are linearly independent)\lambda_i(L)=\min\left(\max_{1\leq j\leq i}\left(\left\lVert v_j\right\rVert\right):v_j\in L\text{ are linearly independent}\right)λi​(L)=min(1≤j≤imax​(∥vj​∥):vj​∈L are linearly independent)

Intuitively, λi(L)\lambda_i(L)λi​(L)is the length of the "ithi^\text{th}ith shortest lattice vector". This intuition is illustrated by the definition of λ1\lambda_1λ1​:

λ1(L)=min⁡(∥v∥:v∈L)\lambda_1(L)=\min\left(\left\lVert v\right\rVert:v\in L\right)λ1​(L)=min(∥v∥:v∈L)

However this is not precise as if vvvis the shortest lattice vector, then −v-v−vis also the shortest lattice vector.

Unfortunately, a basisbib_ibi​for LLLwhere λi(L)=∥bi∥\lambda_i(L)=\left\lVert b_i\right\rVertλi​(L)=∥bi​∥for dimensions 555 and above. This tells us that we can't actually define "the most reduced basis" in contrast to the 2D case (see Lagrange's algorithm) and we would need some other definition to convey this intuition.

An alternate definition ofλi(L)\lambda_i(L)λi​(L)that will be helpful is the radius of the smallest ball centered at the origin such that the ball contains at leastiiilinearly independent vectors inLLL.

Exercises

1) Show that both definitions of λi\lambda_iλi​ are equivalent

2) Consider the lattice L=(2000002000002000002011111)L=\begin{pmatrix}2&0&0&0&0\\0&2&0&0&0\\0&0&2&0&0\\0&0&0&2&0\\1&1&1&1&1\end{pmatrix}L=​20001​02001​00201​00021​00001​​. Show that the successive minima are all222but no basisbib_ibi​can satisfy ∥bi∥=λi\left\lVert b_i\right\rVert=\lambda_i∥bi​∥=λi​.

Minkowski reduced

Definition

The basis{bi}i=1d\left\{b_i\right\}_{i=1}^d{bi​}i=1d​ is Minkowski-reduced if bib_ibi​has minimum length among all vectors in LLL linearly independent from{bj}j=1i−1\left\{b_j\right\}_{j=1}^{i-1}{bj​}j=1i−1​. Equivalently, bib_ibi​has minimum length among all vectors vvvsuch that {b1,…,bi−1,v}\left\{b_1,\dots,b_{i-1},v\right\}{b1​,…,bi−1​,v}can be extended to form a basis ofLLL. Such a notion is strongest among all lattice reduction notions and is generally extremely hard to compute. Another equivalent definition is

∥bi∥≤∥∑j=idcjbj∥gcd⁡(cj)=1\left\lVert b_i\right\rVert\leq\left\lVert\sum_{j=i}^dc_jb_j\right\rVert\quad\gcd\left(c_j\right)=1∥bi​∥≤​j=i∑d​cj​bj​​gcd(cj​)=1

Bounds

λi(L)2≤∥bi∥2≤max⁡(1,(54)i−4)λi(L)2\lambda_i(L)^2\leq\left\lVert b_i\right\rVert^2\leq\max\left(1,\left(\frac54\right)^{i-4}\right)\lambda_i(L)^2λi​(L)2≤∥bi​∥2≤max(1,(45​)i−4)λi​(L)2

The proof presented here is based off [Waerden 1956]. We proceed by induction. Letbib_ibi​be a Minkowski-reduced basis for some latticeLLL. The lower bound is immediate and for i=1i=1i=1, the upper bound is immediate as well.

Let v1,v2…viv_1,v_2\dots v_iv1​,v2​…vi​be linearly independent vectors such that∥vj∥=λj(L)\left\lVert v_j\right\rVert=\lambda_j(L)∥vj​∥=λj​(L). Let Li−1L_{i-1}Li−1​be the sublattice generated by b1,b2,…bi−1b_1,b_2,\dots b_{i-1}b1​,b2​,…bi−1​. Evidently somekkkmust exist such thatvkv_kvk​is not in Li−1L_{i-1}Li−1​. Consider the new lattice L′=L∩span(b1,b2,…bi−1,vk)L'=L\cap\text{span}\left(b_1,b_2,\dots b_{i-1},v_k\right)L′=L∩span(b1​,b2​,…bi−1​,vk​). Letvk′v'_kvk′​be the shortest vector inL′−Li−1L'-L_{i-1}L′−Li−1​such thatb1,b2,…,bi−1,vk′b_1,b_2,\dots,b_{i-1},v'_kb1​,b2​,…,bi−1​,vk′​is a basis for L′L'L′and we have

vk=a1b1+a2b2+⋯+ai−1bi−1+nvk′ai,n∈Z∥bi∥≤∥vk′∥v_k=a_1b_1+a_2b_2+\dots+a_{i-1}b_{i-1}+nv'_k\quad a_i,n\in\mathbb Z\\ \left\lVert b_i\right\rVert\leq\left\lVert v'_k\right\rVertvk​=a1​b1​+a2​b2​+⋯+ai−1​bi−1​+nvk′​ai​,n∈Z∥bi​∥≤∥vk′​∥

Ifn=1n=1n=1, then we are done sincevkv_kvk​can be extended to a basis of LLL, so ∥bi∥≤∥vk∥=λk(L)≤λi(L)\left\lVert b_i\right\rVert\leq\left\lVert v_k\right\rVert=\lambda_k(L)\leq\lambda_i(L)∥bi​∥≤∥vk​∥=λk​(L)≤λi​(L). Otherwise, we have n2≥4n^2\geq4n2≥4. Let vk′=p+qv_k'=p+qvk′​=p+qwherepppis the projection ofvk′v'_kvk′​inLi−1L_{i-1}Li−1​. Since by definition we have∥p∥2≤∥p±bi∥2\left\lVert p\right\rVert^2\leq\left\lVert p\pm b_i\right\rVert^2∥p∥2≤∥p±bi​∥2, we must have

∥p∥2≤14∑j=1i−1∥bj∥2\left\lVert p\right\rVert^2\leq\frac14\sum_{j=1}^{i-1}\left\lVert b_j\right\rVert^2∥p∥2≤41​j=1∑i−1​∥bj​∥2

Furthermore, since

λk2=∥vk∥2=∥a1b1+a2b2+…ai−1bi−1+p∥2+n2∥q∥2\lambda_k^2=\left\lVert v_k\right\rVert^2=\left\lVert a_1b_1+a_2b_2+\dots a_{i-1}b_{i-1}+p\right\rVert^2+n^2\left\lVert q\right\rVert^2λk2​=∥vk​∥2=∥a1​b1​+a2​b2​+…ai−1​bi−1​+p∥2+n2∥q∥2

we have ∥q∥2≤14λk2\left\lVert q\right\rVert^2\leq\frac14\lambda_k^2∥q∥2≤41​λk2​, hence we have

∥bi∥≤14∑j=1i−1∥bj∥2+14λk2≤14∑j=1i−1max⁡(1,(54)i−4)λj(L)2+14λk(L)2≤14(1+∑j=1i−1max⁡(1,(54)i−4))λi(L)2{=max⁡(1,(54)i−4)λi(L)2i≥4<λi(L)2i=2,3\begin{align*} \left\lVert b_i\right\rVert&\leq\frac14\sum_{j=1}^{i-1}\left\lVert b_j\right\rVert^2+\frac14\lambda_k^2\\ &\leq\frac14\sum_{j=1}^{i-1}\max\left(1,\left(\frac54\right)^{i-4}\right)\lambda_j(L)^2+\frac14\lambda_k(L)^2\\ &\leq\frac14\left(1+\sum_{j=1}^{i-1}\max\left(1,\left(\frac54\right)^{i-4}\right)\right)\lambda_i(L)^2\\ &\begin{cases}=\max\left(1,\left(\frac54\right)^{i-4}\right)\lambda_i(L)^2&i\geq4\\<\lambda_i(L)^2&i=2,3\end{cases}\\ \end{align*}∥bi​∥​≤41​j=1∑i−1​∥bj​∥2+41​λk2​≤41​j=1∑i−1​max(1,(45​)i−4)λj​(L)2+41​λk​(L)2≤41​(1+j=1∑i−1​max(1,(45​)i−4))λi​(L)2{=max(1,(45​)i−4)λi​(L)2<λi​(L)2​i≥4i=2,3​​

but since λi(L)2≤∥bi∥2\lambda_i(L)^2\leq \left\lVert b_i\right\rVert^2λi​(L)2≤∥bi​∥2by definition, the case of i=2,3i=2,3i=2,3cannot occur here (hence n=1n=1n=1in these cases).

Exercises

1) Show that both definitions of Minkowski-reduced lattice are equivalent

2) Consider the lattice L=(2000002000002000002011111)L=\begin{pmatrix}2&0&0&0&0\\0&2&0&0&0\\0&0&2&0&0\\0&0&0&2&0\\1&1&1&1&1\end{pmatrix}L=​20001​02001​00201​00021​00001​​. We have showed in a previous exercise that the successive minima are all222but no basisbib_ibi​can satisfy ∥bi∥=λi\left\lVert b_i\right\rVert=\lambda_i∥bi​∥=λi​, show that for any Minkowski reduced basis bib_ibi​, the basis must satisfy ∥bi∥2=max⁡(1,(54)i−4)λi(L)2\left\lVert b_i\right\rVert^2=\max\left(1,\left(\frac54\right)^{i-4}\right)\lambda_i(L)^2∥bi​∥2=max(1,(45​)i−4)λi​(L)2

HKZ reduced

Definition

Let πi\pi_iπi​as the projection to the orthogonal complement of {bj}j=1i−1\left\{b_j\right\}_{j=1}^{i-1}{bj​}j=1i−1​.Then the basis is HKZ-reduced if it is size-reduced and ∣∣bi∗∣∣=λ1(πi(L))||b_i^*||=\lambda_1\left(\pi_i(L)\right)∣∣bi∗​∣∣=λ1​(πi​(L)). This definition gives us a relatively simple way to compute a HKZ-reduced basis by iteratively finding the shortest vector in orthogonal projections.

Bounds

4i+3≤(∣∣bi∣∣λi(L))2≤i+34\frac4{i+3}\leq\left(\frac{||b_i||}{\lambda_i(L)}\right)^2\leq\frac{i+3}4i+34​≤(λi​(L)∣∣bi​∣∣​)2≤4i+3​

LLL reduced

Definition

Let δ∈(14,1)\delta\in\left(\frac14,1\right)δ∈(41​,1). A basis{bi}i=1d\left\{b_i\right\}_{i=1}^d{bi​}i=1d​is δ\deltaδ- LLL-reduced if it is size reduced and satisfy the Lovász condition, i.e.

δ∥bi∗∥2≤∥bi+1∗+μi+1,ibi∗∥2\delta\left\lVert b_i^*\right\rVert^2\leq\left\lVert b_{i+1}^*+\mu_{i+1,i}b_i^*\right\rVert^2δ∥bi∗​∥2≤​bi+1∗​+μi+1,i​bi∗​​2

This notion of reduction is most useful to use for fast algorithms as such a basis can be found in polynomial time (see LLL reduction).

Bounds

∥b1∥≤(44δ−1)d−14vol(L)1d∥bi∥≤(44δ−1)d−12λi(L)∏i=1d∥bi∥≤(44δ−1)d(d−1)4vol(L)\begin{align*} \left\lVert b_1\right\rVert&\leq\left(\frac4{4\delta-1}\right)^{\frac{d-1}4}\text{vol}(L)^\frac1d\\ \left\lVert b_i\right\rVert&\leq\left(\frac4{4\delta-1}\right)^{\frac{d-1}2}\lambda_i(L)\\ \prod_{i=1}^d\left\lVert b_i\right\rVert&\leq\left(\frac4{4\delta-1}\right)^{\frac{d(d-1)}4}\text{vol}(L) \end{align*}∥b1​∥∥bi​∥i=1∏d​∥bi​∥​≤(4δ−14​)4d−1​vol(L)d1​≤(4δ−14​)2d−1​λi​(L)≤(4δ−14​)4d(d−1)​vol(L)​